(a) x+a=0 and b-x=0 are the solutions/roots to f(x)=0, i.e. where the graph cuts the x-axis. Notice that the intercept at x=-4 is a turning point (gradient to the left is negative, gradient to the right is positive), hence that root is repeated, so it must be squared in f(x). Hence x+4=0, a=4. The other intercept at x=3 must be the other solution, 3-x=0, so b=3.

(b) Translate the entire graph 2 units in the negative x-direction, i.e. move the graph 2 units to the left. So the x-intercepts become (-6,0) and (1,0). The y=intercept (0,48) becomes (-2,48) (not y-intercept anymore and you must lable that). We don’t know what the new y-intercept is yet, so we don’t need to lable it. Note that the “shape” of the graph must be the same.

(c) First substitute all the x in f(x) with (x+2).

equation of curve C: y=f(x+2)

=(x+2+4)²(3-(x+2))

=(x+6)²(1-x)

=(x²+12x+36)(1-x)

=x²+12x+36-x³-12x²-36x

so equation of C is y= -x³-11x²-24x+36

p=-1, q=-11, r=-24, s=36

(d) We simply sub x=0 into the equation above (since we’re finding the y=intercept)

y=36

so C meets the y-axis at (0,36).